Imaginary Powers

Algebra Level 4

( − 1 ) − i i i = ( a a ) π , a = ? \large \frac { (-1)^{-i}}{i^i} = (a \sqrt a)^\pi, \ \ \ \ \ a = \ ?

Give your answer to 3 decimal places. Note that i = − 1 i = \sqrt{-1} .


The answer is 2.718.

Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Michael Mendrin
Apr 26, 2015

First, we start with Euler’s famous identity

e π i = − 1 { e }^{ \pi i }=-1

From this, we can work out the expression on the left side

( − 1 ) − i i i = ( e π i ) − i ( e 1 2 π i ) i = e π e − 1 2 π = e 3 2 π \dfrac { { \left( -1 \right) }^{- i } }{ { i }^{ i } } =\dfrac { { \left( { e }^{ \pi i } \right) }^{ -i } }{ { \left( { e }^{ \frac { 1 }{ 2 } \pi i } \right) }^{ i } } =\dfrac { { e }^{ \pi } }{ { e }^{ -\frac { 1 }{ 2 } \pi } } ={ e }^{ \frac { 3 }{ 2 } \pi }

From there, it’s a short step to seeing that a = e a=e

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Great! Can you find the flaw here?

( − 1 ) − i i i = ( ( − 1 ) − 1 i ) i \LARGE \frac { (-1)^{-i}}{i^i} = \left ( \frac {(-1)^{-1} }{i} \right )^i

= ( − 1 i ) i = i i = e − π / 2 = a 3 π / 2 = \left ( \frac {-1 }{i} \right )^i = i ^i = e^{-\pi /2} = a^{3\pi /2} .

Then a = e − 1 / 3 a = e^{-1/3} .

2nd Method:

( ( − 1 ) − i i i ) = ( ( − 1 ) − 1 i i ) i = ( − i ) i \LARGE \left ( \frac { (-1)^{-i}}{i^i} \right ) = \left ( \frac { (-1)^{-1}}{i^i} \right )^i = (-i)^i

Raised both sides to the power of i i :

( ( − i ) i ) i = ( ( a a ) π ) i = a 3 π i / 2 ( (-i)^i )^i = ((a \sqrt a)^\pi )^i = a^{3\pi i /2}

( − i ) i 2 = a 3 π i / 2 (-i)^{i^2} = a^{3\pi i /2} , then ( − i ) − 1 = − i = a 3 π i / 2 (-i)^{-1} = -i = a^{3\pi i /2}

Square both sides

− 1 = a 3 π i -1 = a^{3\pi i} , then e i π = a 3 π i e^{i \pi } = a^{3\pi i} , so a 3 = e a^3 = e or a = e 1 / 3 a = e^{1/3} .

Continued on this note

We know that

( − 1 ) i = i 2 i = 0.0432139... { \left( -1 \right) }^{ i }={ i }^{ 2i }=0.0432139...

Now, but we also know that

( − 1 ) 3 = ( − 1 ) { \left( -1 \right) }^{ 3 }=\left( -1 \right)

So that

( − 1 ) 3 i = ( − 1 ) i { \left( -1 \right) }^{ 3i }={ \left( -1 \right) }^{ i }

Which means that

( 0.0432139... ) 3 = 0.0432139... ? { \left( 0.0432139... \right) }^{ 3 }=0.0432139...?\quad

The fallacy comes from the fact that when − 1 -1 is raised to the i i power, it results in a number which is not − 1 -1 , 0 0 or 1 1 , so we have to be careful about inadvertently introducing such factors in this fashion or in similar ways. I think all of the fallacious examples given above can be explained by such inadvertent factors being introduced. Whenever we have a complex power of a negative number, be very, very careful what you do with it. The derivation I used here avoids this trap.

I agree that for a Level 3 problem, this problem contains some really tricky pitfalls involved in doing computations with i i .

Michael Mendrin - 6 years, 1 month ago

In both cases the flow is that you are using a radical property that can't be used always in complex numbers, just like in this problem

Hjalmar Orellana Soto - 5 years, 9 months ago

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